一个三角形不等式
先给几个熟知结论，S为面积
S^2=s(s-a)(s-b)(s-c), S=abc/(4R)=sr
于是(s-a)(s-b)(s-c)=sr^2
于是有abc+(s-a)(s-b)(s-c)=4sRr+sr^2
=s^3+s(ab+bc+ca)
ab+bc+ca=s^2+4Rr+r^2,a^2+b^2+c^2=2(s^2-2Rr-r^2)
要证楼上所说的Gerresten不等式：4R^2+4Rr+3r^2>=s^2
记x=s-a,y=s-b,z=s-c
p=x^2y+y^2z+z^2x+zx^2+xy^2+yz^2,q=xyz
(4R^2+4Rr+3r^2-s^2)*4r^2s^2=(a...全部
先给几个熟知结论，S为面积
S^2=s(s-a)(s-b)(s-c), S=abc/(4R)=sr
于是(s-a)(s-b)(s-c)=sr^2
于是有abc+(s-a)(s-b)(s-c)=4sRr+sr^2
=s^3+s(ab+bc+ca)
ab+bc+ca=s^2+4Rr+r^2,a^2+b^2+c^2=2(s^2-2Rr-r^2)
要证楼上所说的Gerresten不等式：4R^2+4Rr+3r^2>=s^2
记x=s-a,y=s-b,z=s-c
p=x^2y+y^2z+z^2x+zx^2+xy^2+yz^2,q=xyz
(4R^2+4Rr+3r^2-s^2)*4r^2s^2=(abc)^2+4abc(s-a)(s-b)(s-c)+
12[(s-a)(s-b)(s-c)]^2-4s^3(s-a)(s-b)(s-c)
=[(x+y)(y+z)(z+x)]^2+4xyz(x+y)(y+z)(z+x)+12(xyz)^2
-4(x+y+z)^2xyz
=(p+2q)^2+4(p+2q)q+12q^2-4q(x^3+y^3+z^3+3p+6q)
=p^2-4q(x^3+y^3+z^3)-4pq
=x^4(y+z)^2+y^4(z+x)^2+z^4(x+y)^2+2∑x^2y^2(y+z)(x+z)-4q(x^3+y^3+z^3)-4pq
≥4xyz(x^3+y^3+z^3)+2∑x^2y^2(y+z)(x+z)-4q(x^3+y^3+z^3)
-4pq
=2∑x^2y^2(y+z)(x+z)-4pq
=6x^2y^2z^2+2(xy+yz+zx)(x^2y^2+y^2z^2+z^2x^2)-4pq
=2∑(x^2y^2z^2-x^3y^2z-x^2y^3z+x^3y^3)
=2∑x^2y^2(x-z)(y-z)
不妨设x≥y≥z
∑x^2y^2(x-z)(y-z)=x^2y^2(x-z)(y-z)+y^2z^2(x-y)(x-z)+
-z^2x^2(x-y)(y-z)
=x^2y^2(x-y+y-z)(y-z)+y^2z^2(x-y)(x-y+y-z)
-z^2x^2(x-y)(y-z)
=(x^2y^2+y^2z^2-z^2x^2)(x-y)(y-z)+x^2y^2(y-z)^2
+y^2z^2(x-y)^2
≥0
所以不等式得证！。
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